Odpowiedź:
[tex]2\sqrt{6}(\sqrt{6}+2x)-3x(x^2-3\sqrt{6})=2\cdot6+4\sqrt{6}x-3x^3+9\sqrt{6}x=12+13\sqrt{6}x-3x^3\\\\\\dla\ \ x=-\sqrt{6}\\\\12+13\sqrt{6}x-3x^3=12+13\sqrt{6}\cdot(-\sqrt{6})-3\cdot(-\sqrt{6})^3=12-13\cdot6-3\cdot(-\sqrt{216})=\\\\=12-78+3\sqrt{216}=-66+3\sqrt{36\cdot6}=-66+3\cdot6\sqrt{6}=-66+18\sqrt{6}\\\\Odp.Warto\'s\'c\ \ wyra\.zenia\ \ jest\ \ r\'owna\ \ 18\sqrt{6}+(-66)[/tex]
